MOMENT OF INERTIA
Moments of Area
Example 1: Pressure on a dam
The left figure shows a dam bounding a body of water. Acting at
any point is a water pressure which is a force per unit of area
Water pressure is a function of both water specific weight ( γ ) and
water depth (x). The pressure is constant across an area of
infinitely small height dx and width 'L' but will vary linearly with
height 'x' as p  x =⋅x
2
First Moment of Area
Assume pressure p(x) is applied to
elemental area dA.
Element dA is located a distance 'x' below
the z-axis.
If we multiply this elemental area by distance
'x', we can define elemental moment:
dM = x⋅dA
or if expressed algebraically
M i = x⋅Ai
Integrating, we find the statical or first
moment of area as: M = ∫A x⋅dA
expressed algebraically as
M = ∑ x⋅A i
i
3
Second Moment of Area
If we take an element of area dA and
multiply it by pressure p(x)...
...the elemental force acting on dA is:
dF = p  x⋅dA
If we multiply this elemental moment by its
distance 'x' from the z-axis, we have:
dF = p  x⋅dA
The moment of dF about the z-axis is:
dM = x⋅dF
2
dM = x⋅ ⋅x⋅dA = ⋅x ⋅dA
4
Second Moment of Area
If we define γ as a constant and divide it out, we are left with:
dM = x⋅ x⋅dA
Mathematically, this is a moment of the first moment of area,
thus we refer to this as the second moment of area.
The moment of the entire distribution is equal to the integral
of dM over the entire area.
2
M = ∫A dM = ∫A x dA
Algebraically, we can replace elemental area dA with some
area Ai of finite size. Total moment is then the sum of all
individual moments.
M = ∑  x 2⋅Ai 
i
5
Second Moment of Area
Thus, the second moment of area...
...is often called area moment of inertia due to the
similarity to the mathematical form of the integral for the
resultant moments of the inertia forces in the case of
rotating bodies (i.e.: mass moment of inertia).
For our purposes, we will hereafter refer to the second
moment of area simply as the moment of inertia
As such, the moment of interia is defined as:
2
I = ∫A distance dA
An algebraic approximation can be expressed as:
I = ∑ distance 2 Ai
i
6
Area Moment of Inertia
Example 2:
Beam Bending - Stress
A common application is to
determine the internal bending
stress in a beam. Such stresses
are linear.
Analysis of these stresses is a
strength of materials topic and
does require the moment of
inertia of the beam cross section.
This will be addressed in detail in
your strength of materials course
y
b =
7
M⋅y
I
Area Moment of Inertia
Example 3:
Beam Bending – Deflection
Consider bending the same
beam in two planes...for
example, a 2 x 4 laid across a
pair of carpenter's horses.
Now try with your own ruler.
About which axis is the ruler
'stiffer'?
Study the equation for moment of inertia again.
2
I = ∫A distance dA
Notice that for a given cross section, it is not area that matters,
it is the orientation of the section.
8
Area Moment of Inertia
As stated previously, the second moment of area, or area
moment of inertia, is given by:
I = ∫A distance 2 dA
Let's consider finding the moment of inertia about some
arbitrary 'x' or 'y' axis. Consider cross section below with area
'A'. Elemental area dA is located a
Y
x
distance 'x' from the y-axis and a
distance 'y' from the x-axis. Then:
I x = ∫A y 2 dA
I y = ∫A x 2 dA
dA
y
X
The units of moment of inertia will be length to the fourth power
such as in4, ft4, mm4, or m4
9
Centroidal Moment of Inertia
In most cases, we are interested in the moment of inertia
about an x-y axis superimposed on the centroid of the cross
section. This is known as the centroidal moment of inertia.
Centroidal moments of inertia for simple geometric shapes
can be found in many engineering handbooks, on the Internet,
and in the appendix of this online text.
But before we use such handbook data, let's apply the
general equation for moment of inertia to a rectangular cross
section.
10
Centroidal Moment of Inertia
Consider a rectangle with a base width 'b' and a height 'h'.
The application of calculus will yield:
y
dA = b⋅dy
dy
h
I xc = ∫ y ⋅dA =
A
∫
2
y b⋅dy
−h / 2
[    ]
3
3
y
c
h /2
2
h/ 2
y
b h
−h
= b⋅ ∣ =
−
3 −h/ 2 3 2
2
x
3
=
b⋅h
12
3
y
x dx
b
b/ 2
I yc = ∫ x ⋅dA =
2
A
∫
2
c
x
−b/ 2
3
[    ]
3
b /2
x
h b
−b
= h⋅ ∣ =
−
3 −b / 2 3 2
2
11
h
x  h⋅dx
3
=
h⋅b
12
dA = h⋅dx
3
b
Centroidal Moment of Inertia
We can also estimate the moment of inertia using algebraic
methods. By breaking the area into finite sections (lets try
four), then the moment of inertia (Ixc only) calculates as:
y
Ai = b⋅
h/ 4
3h
8
h
3h
8
h/ 4
h
8
h
8
h/ 4
h/ 4
b
x
h
4
2
I xc =  y ⋅A i
[  ] [  ] [  ] [   ]
[        ]
2
2
2
2
3h
h
h
h
h
h
3h
h
=
⋅b⋅ 
⋅b⋅ 
⋅b⋅ 
⋅b⋅
8
4
8
4
8
4
8
4
2
3h
=
8
5bh
=
64
h

8
2
h

8
2
3h

8
2
h
⋅b⋅
4
3
Note this does not equal the moment of inertia calculated using
calculus methods. We can get closer by breaking the area into
more and smaller rectangles.
12
Centroidal Moment of Inertia
For example, by dividing the rectangle into 8 equal areas, we
would calculate the moment of inertia to be:
I xc =  y 2⋅Ai
[                ]
7h
=
16

2
5h

16
2

2
3h

16
168 h
h
=
⋅b⋅ =
256
8
2
21bh
256
h

16
2
h

16
2
3h

16
2
5h

16
2
7h

16
2
h
⋅b⋅
8
3
Although much closer to the result obtained through
integration, it is still not equal. Although we could continue
to divide this rectangle into even smaller areas to refine the
solution, it is unneccessary. Integrations have been
accomplished for common geometric shapes and are
tabulated in the appendix to this online text.
13
Centroidal Moment of Inertia
When applying the relationships listed in the appendix, take
care to note the location of the x-y coordinate axis. Moments
of inertia can be found about any axis one wishes.
For example, the moment of inertia of a rectangle about an
x-axis depends on the x-axis in which you are interested, the
centroidal axis or the base axis.
y
h
y'
x'
c
b
14
x
b⋅h 3
I x' =
12
b⋅h3
Ix=
3
b 3⋅h
I y' =
12
3
b ⋅h
Iy =
3